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11_ch 17 Mechanical Design budynas_SM_ch17

# 11_ch 17 Mechanical Design budynas_SM_ch17 - 916 3 ft/s S y...

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430 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design Drive geometry: d = 2 in, D = 4 in Belt thickness: t = 0 . 003 in Design variables: Belt width b Belt loop periphery Preliminaries H d = H nom K s n d = 1(1 . 2)(1 . 05) = 1 . 26 hp T = 63 025(1 . 26) 1750 = 45 . 38 lbf · in A 15 in center-to-center distance corresponds to a belt loop periphery of 39.5 in. The 40 in loop available corresponds to a 15.254 in center distance. θ d = π 2 sin 1 4 2 2(15 . 254) = 3 . 010 rad θ D = π + 2 sin 1 4 2 2(15 . 274) = 3 . 273 rad For full friction development exp( f θ d ) = exp[0 . 35(3 . 010)] = 2 . 868 V = π dn 12 = π (2)(1750)
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Unformatted text preview: 916 . 3 ft/s S y = 175 000 psi Eq. (17-15): S f = 14 . 17(10 6 )(10 6 ) − . 407 = 51 212 psi From selection step 3 a = ± S f − Et (1 − ν 2 ) d ² t = ± 51 212 − 28(10 6 )(0 . 003) (1 − . 285 2 )(2) ² (0 . 003) = 16 . 50 lbf/in of belt width ( F 1 ) a = ab = 16 . 50 b For full friction development, from Prob. 17-13, b min = ± F a exp( f θ d ) exp( f θ d ) − 1 ± F = 2 T d = 2(45 . 38) 2 = 45 . 38 lbf So b min = 45 . 38 16 . 50 ³ 2 . 868 2 . 868 − 1 ´ = 4 . 23 in...
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